3.576 \(\int e^{2 \tanh ^{-1}(a x)} \sqrt {c-\frac {c}{a x}} \, dx\)

Optimal. Leaf size=51 \[ x \left (-\sqrt {c-\frac {c}{a x}}\right )-\frac {3 \sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {c-\frac {c}{a x}}}{\sqrt {c}}\right )}{a} \]

[Out]

-3*arctanh((c-c/a/x)^(1/2)/c^(1/2))*c^(1/2)/a-x*(c-c/a/x)^(1/2)

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Rubi [A]  time = 0.10, antiderivative size = 51, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.292, Rules used = {6133, 25, 514, 375, 78, 63, 208} \[ x \left (-\sqrt {c-\frac {c}{a x}}\right )-\frac {3 \sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {c-\frac {c}{a x}}}{\sqrt {c}}\right )}{a} \]

Antiderivative was successfully verified.

[In]

Int[E^(2*ArcTanh[a*x])*Sqrt[c - c/(a*x)],x]

[Out]

-(Sqrt[c - c/(a*x)]*x) - (3*Sqrt[c]*ArcTanh[Sqrt[c - c/(a*x)]/Sqrt[c]])/a

Rule 25

Int[(u_.)*((a_) + (b_.)*(x_)^(n_.))^(m_.)*((c_) + (d_.)*(x_)^(q_.))^(p_.), x_Symbol] :> Dist[(d/a)^p, Int[(u*(
a + b*x^n)^(m + p))/x^(n*p), x], x] /; FreeQ[{a, b, c, d, m, n}, x] && EqQ[q, -n] && IntegerQ[p] && EqQ[a*c -
b*d, 0] &&  !(IntegerQ[m] && NegQ[n])

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 78

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> -Simp[((b*e - a*f
)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(f*(p + 1)*(c*f - d*e)), x] - Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1)
+ c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)), Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f,
n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || IntegerQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || LtQ
[p, n]))))

Rule 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rule 375

Int[((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> -Subst[Int[((a + b/x^n)^p*(c +
d/x^n)^q)/x^2, x], x, 1/x] /; FreeQ[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && ILtQ[n, 0]

Rule 514

Int[(x_)^(m_.)*((c_) + (d_.)*(x_)^(mn_.))^(q_.)*((a_) + (b_.)*(x_)^(n_.))^(p_.), x_Symbol] :> Int[x^(m - n*q)*
(a + b*x^n)^p*(d + c*x^n)^q, x] /; FreeQ[{a, b, c, d, m, n, p}, x] && EqQ[mn, -n] && IntegerQ[q] && (PosQ[n] |
|  !IntegerQ[p])

Rule 6133

Int[E^(ArcTanh[(a_.)*(x_)]*(n_))*(u_.)*((c_) + (d_.)/(x_))^(p_), x_Symbol] :> Int[(u*(c + d/x)^p*(1 + a*x)^(n/
2))/(1 - a*x)^(n/2), x] /; FreeQ[{a, c, d, p}, x] && EqQ[c^2 - a^2*d^2, 0] &&  !IntegerQ[p] && IntegerQ[n/2] &
&  !GtQ[c, 0]

Rubi steps

\begin {align*} \int e^{2 \tanh ^{-1}(a x)} \sqrt {c-\frac {c}{a x}} \, dx &=\int \frac {\sqrt {c-\frac {c}{a x}} (1+a x)}{1-a x} \, dx\\ &=-\frac {c \int \frac {1+a x}{\sqrt {c-\frac {c}{a x}} x} \, dx}{a}\\ &=-\frac {c \int \frac {a+\frac {1}{x}}{\sqrt {c-\frac {c}{a x}}} \, dx}{a}\\ &=\frac {c \operatorname {Subst}\left (\int \frac {a+x}{x^2 \sqrt {c-\frac {c x}{a}}} \, dx,x,\frac {1}{x}\right )}{a}\\ &=-\sqrt {c-\frac {c}{a x}} x+\frac {(3 c) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {c-\frac {c x}{a}}} \, dx,x,\frac {1}{x}\right )}{2 a}\\ &=-\sqrt {c-\frac {c}{a x}} x-3 \operatorname {Subst}\left (\int \frac {1}{a-\frac {a x^2}{c}} \, dx,x,\sqrt {c-\frac {c}{a x}}\right )\\ &=-\sqrt {c-\frac {c}{a x}} x-\frac {3 \sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {c-\frac {c}{a x}}}{\sqrt {c}}\right )}{a}\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 51, normalized size = 1.00 \[ x \left (-\sqrt {c-\frac {c}{a x}}\right )-\frac {3 \sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {c-\frac {c}{a x}}}{\sqrt {c}}\right )}{a} \]

Antiderivative was successfully verified.

[In]

Integrate[E^(2*ArcTanh[a*x])*Sqrt[c - c/(a*x)],x]

[Out]

-(Sqrt[c - c/(a*x)]*x) - (3*Sqrt[c]*ArcTanh[Sqrt[c - c/(a*x)]/Sqrt[c]])/a

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fricas [A]  time = 0.53, size = 125, normalized size = 2.45 \[ \left [-\frac {2 \, a x \sqrt {\frac {a c x - c}{a x}} - 3 \, \sqrt {c} \log \left (-2 \, a c x + 2 \, a \sqrt {c} x \sqrt {\frac {a c x - c}{a x}} + c\right )}{2 \, a}, -\frac {a x \sqrt {\frac {a c x - c}{a x}} - 3 \, \sqrt {-c} \arctan \left (\frac {\sqrt {-c} \sqrt {\frac {a c x - c}{a x}}}{c}\right )}{a}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)^2/(-a^2*x^2+1)*(c-c/a/x)^(1/2),x, algorithm="fricas")

[Out]

[-1/2*(2*a*x*sqrt((a*c*x - c)/(a*x)) - 3*sqrt(c)*log(-2*a*c*x + 2*a*sqrt(c)*x*sqrt((a*c*x - c)/(a*x)) + c))/a,
-(a*x*sqrt((a*c*x - c)/(a*x)) - 3*sqrt(-c)*arctan(sqrt(-c)*sqrt((a*c*x - c)/(a*x))/c))/a]

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giac [B]  time = 0.23, size = 97, normalized size = 1.90 \[ -\frac {3 \, \sqrt {c} \log \left ({\left | a \right |} {\left | c \right |}\right ) \mathrm {sgn}\relax (x)}{2 \, a} + \frac {3 \, \sqrt {c} \log \left ({\left | -2 \, {\left (\sqrt {a^{2} c} x - \sqrt {a^{2} c x^{2} - a c x}\right )} \sqrt {c} {\left | a \right |} + a c \right |}\right )}{2 \, a \mathrm {sgn}\relax (x)} - \frac {\sqrt {a^{2} c x^{2} - a c x} {\left | a \right |}}{a^{2} \mathrm {sgn}\relax (x)} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)^2/(-a^2*x^2+1)*(c-c/a/x)^(1/2),x, algorithm="giac")

[Out]

-3/2*sqrt(c)*log(abs(a)*abs(c))*sgn(x)/a + 3/2*sqrt(c)*log(abs(-2*(sqrt(a^2*c)*x - sqrt(a^2*c*x^2 - a*c*x))*sq
rt(c)*abs(a) + a*c))/(a*sgn(x)) - sqrt(a^2*c*x^2 - a*c*x)*abs(a)/(a^2*sgn(x))

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maple [B]  time = 0.04, size = 120, normalized size = 2.35 \[ \frac {\sqrt {\frac {c \left (a x -1\right )}{a x}}\, x \left (2 \sqrt {a \,x^{2}-x}\, \sqrt {a}-4 \sqrt {\left (a x -1\right ) x}\, \sqrt {a}-\ln \left (\frac {2 \sqrt {a \,x^{2}-x}\, \sqrt {a}+2 a x -1}{2 \sqrt {a}}\right )-2 \ln \left (\frac {2 \sqrt {\left (a x -1\right ) x}\, \sqrt {a}+2 a x -1}{2 \sqrt {a}}\right )\right )}{2 \sqrt {\left (a x -1\right ) x}\, \sqrt {a}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*x+1)^2/(-a^2*x^2+1)*(c-c/a/x)^(1/2),x)

[Out]

1/2*(c*(a*x-1)/a/x)^(1/2)*x*(2*(a*x^2-x)^(1/2)*a^(1/2)-4*((a*x-1)*x)^(1/2)*a^(1/2)-ln(1/2*(2*(a*x^2-x)^(1/2)*a
^(1/2)+2*a*x-1)/a^(1/2))-2*ln(1/2*(2*((a*x-1)*x)^(1/2)*a^(1/2)+2*a*x-1)/a^(1/2)))/((a*x-1)*x)^(1/2)/a^(1/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -\int \frac {{\left (a x + 1\right )}^{2} \sqrt {c - \frac {c}{a x}}}{a^{2} x^{2} - 1}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)^2/(-a^2*x^2+1)*(c-c/a/x)^(1/2),x, algorithm="maxima")

[Out]

-integrate((a*x + 1)^2*sqrt(c - c/(a*x))/(a^2*x^2 - 1), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int -\frac {\sqrt {c-\frac {c}{a\,x}}\,{\left (a\,x+1\right )}^2}{a^2\,x^2-1} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-((c - c/(a*x))^(1/2)*(a*x + 1)^2)/(a^2*x^2 - 1),x)

[Out]

int(-((c - c/(a*x))^(1/2)*(a*x + 1)^2)/(a^2*x^2 - 1), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - \int \frac {\sqrt {c - \frac {c}{a x}}}{a x - 1}\, dx - \int \frac {a x \sqrt {c - \frac {c}{a x}}}{a x - 1}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)**2/(-a**2*x**2+1)*(c-c/a/x)**(1/2),x)

[Out]

-Integral(sqrt(c - c/(a*x))/(a*x - 1), x) - Integral(a*x*sqrt(c - c/(a*x))/(a*x - 1), x)

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